What is the sum from n=1 to infinity of 4/(n!) ?

The sum from n=1 to infinity of 4/(n!) is converges

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sum from n=1 to infinity of 4/(n!)

n = 1 \sum \infty \frac{4}{n !}

converges

Solution steps

n = 1 \sum \infty \frac{4}{n !}

Apply the constant multiplication rule:

\sum c \cdot a_{n} = c \cdot \sum a_{n}

= 4 \cdot n = 1 \sum \infty \frac{1}{n !}

Apply Series Ratio Test:

converges

= 4 converges

= converges

n = 1 \sum \infty \frac{n ^{9}}{9 ^{n}}

n = 1 \sum \infty n^{4} e^{- n^{2}}

n = 0 \sum \infty (\frac{j}{1 + j})^{n}

n = 1 \sum \infty \frac{e ^{\frac{4}{n}}}{n}

n = 1 \sum \infty \frac{e ^{\frac{2}{n}}}{n}

What is the sum from n=1 to infinity of 4/(n!) ?
The sum from n=1 to infinity of 4/(n!) is converges